Nearly Tight Linear Programming Bounds for Demand Matching in Bipartite Graphs

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1 Nearly Tight Linear Programming Bounds for Demand Matching in Biartite Grahs Mohit Singh Hehui Wu Setember 23, 212 Abstract We consider the demand matching roblem which is a generalization of the maximum weight biartite matching roblem as well as the b-matching roblem We are given a simle biartite grah G = (V, E), a demand function d and a rofit function π on edges and a caacity function b on vertices A subset M of edges is called a demand matching if the sum of demands d e of edges chosen in M incident at v is at most b v for each vertex v The goal of the demand matching roblem is to select a demand matching M which maximizes the sum of rofit of edges in M When all demands d e = 1, this roblem is exactly the b-matching roblem In this aer we give nearly tight uer and lower bounds on the integrality ga of a natural linear rogramming relaxation for the roblem Our first result is to show that the integrality ga is bounded from above by the fractional coloring number of a tree-net A tree-net is a grah obtained by connecting non-adjacent vertices of a tree by vertex disjoint aths of length at least two We then give an exlicit bound of 279 on the fractional chromatic number of any tree-net which also results in a 279-aroximation algorithm To comlement this algorithm, we exlicitly show a lower bound of 2699 on the integrality ga by constructing tree-net grahs whose fractional chromatic number is at least 2699 Microsoft Research, Redmond and McGill University McGill University 1

2 1 Introduction Maximum weight matching roblem in biartite grahs is a cornerstone roblem in combinatorial otimization with efficient algorithms given by Kuhn [Kuh55] building on the wor of König [Kön31] and Egerváry [Eug31] The roblem finds many alications and various generalizations have been studied, including the b-matching roblem In this aer, we study the demand matching roblem, introduced by Sheherd and Vetta [SV7], which combines the asects of nasac roblem along with biartite matching roblem In an instance of demand matching roblem we are given a simle biartite grah G = (U V, E), a rofit function π : E R +, a demand function d : E R + and a caacity function on vertices c : V R + A demand matching is defined as a subgrah M E such that e δ(v) M d e b v for each v U V The goal of the demand matching roblem is to find a demand matching which maximizes rofit, e M π e If d e = 1 for each e E and b v = 1 for each v U V then it is the standard maximum weight matching and for arbitrary b, it is exactly the b-matching roblem The roblem can also be interreted as a collection of nasac roblems, one at each vertex v, which are couled since edges occur as items in two distinct nasac roblem The notion of introducing demands and the all or nothing constraint has been successfully alied to many roblems The generalized assignment roblem [ST93] where the goal is to assign jobs to machines with arbitrary job sizes is closely related to the demand matching roblem In this roblem, the constraint at the vertices, corresonding to maesan on machines, is a soft constraint unlie in the demand matching roblem where the caacity constraint at the vertices is a hard constraint A further generalization is the unslittable flow roblem studied by Kleinberg [Kle96] where the goal is to route flow aths between airs of vertices in grah satisfying certain caacities and each air has a certain demand For each air, either we have to route the whole demand or none of it Another related roblem arises from the design of communication switches called Clos networs [Clo53] Given the same inut as in the demand matching roblem, the goal here is to find a coloring of edges with the number of colors where each color class is a demand matching Most wor on this roblem has been focused on instances when b v = 1 for each vertex v V 11 Our Results and Techniques We study the standard linear rogramming relaxation for the demand matching roblem given in Figure 1 Our first result is an aroximation algorithm which gives an uer bound on the integrality ga of the linear roblem Theorem 1 (Uer Bound) There is a randomized algorithm which returns a demand matching of exected rofit at least times the rofit of the otimal solution to LP dem The main technique behind the algorithm is to reduce the roblem to fractionally coloring a tree-net A grah G is a tree-net if G is triangle free and G \ V 2 is acyclic where V 2 is the set of vertices of degree two It is equivalent to the fact that G can be obtained by connecting non-adjacent vertices of a tree with vertex disjoint aths of length at least two We rove the following lemma which relates the integrality ga to the fractional coloring number χ f (G) of a tree net G The fractional coloring number of a grah G is the minimum number of indeendent sets needed to obtain a fractional covering of the vertices 2

3 Lemma 2 If for each tree-net G we have χ f (G) ρ, then the integrality ga of the demand matching linear rogram is at most ρ Next we design an algorithm which finds a good fractional coloring for any tree-net The following lemma also gives the algorithm that comletes the roof of Theorem 1 Lemma 3 χ f (G) 279 for every tree-net G To comlement our aroximation algorithm we rove the following lower bound on the integrality ga Theorem 4 (Lower Bound) There exists an instance of the demand matching roblem such that the integrality ga is at least 2699 Interestingly, the construction of the integrality ga also follows via tree-nets We construct a tree-net whose fractional chromatic number is at least 2699 Moreover, we also ensure that the tree-net satisfies certain roerties that enables us to trace bac the reduction done for the aroximation algorithm and obtain an instance of the demand matching exhibiting the integrality ga 12 Related Wor As mentioned above, the demand matching roblem was first introduced by Sheherd and Vetta [SV7] who gave uer and lower bounds on the integrality ga of linear rogram They showed integrality ga for biartite grahs is in the interval [25, 276] and for general grahs in the interval [3, 3264] Pareh [Par11] later showed that the integrality ga on general grahs is exactly 3 Multile nasac roblem was studied by Cheuri and Khanna [CK5] who gave a olynomial time aroximation scheme for the roblem For the generalized assignment roblem, Shmoys and Tardos [ST93] gave an algorithm which returns an assignment of otimal cost and achieves a 2-aroximation for maesan Kleinberg [Kle96] studied the single source unslittable flow roblem for which tight bounds were shown by Dinitz, Garg and Goemans [DGG99] The version with costs on the edges has been studied by Sutella [KS1, Su] Clos networs were introduced by Clos [Clo53] to obtain better communication networs The underlying combinatorial roblem of coloring a weighted biartite grah where each color class is a demand matching has received considerable interest [NV3, DGHK99, CG7, FS8] Our results also imly a fractional coloring for the general case with arbitrary b v while the attention in Clos networs has been to the case when b v = 1 for each v V 2 Aroximation Algorithm for Demand Matching In this section, we will rove Theorem 1 by giving an aroximation algorithm First we reduce the roblem of finding an aroximation algorithm to fractionally coloring a tree-net Later we give an algorithm to construct such a coloring 21 Reduction to Coloring Tree-nets In this section, we rove Lemma 2 First we give a few definitions The fractional chromatic number of a grah G is defined to be otimum solution to the following linear rogram Here I denotes the set of all indeendent sets of G The otimum solution to the linear rogram gives the fractional coloring number χ f (G) for any grah G 3

4 (IP ) max e E π ex e (LP dem ) max e E π ex e : : v V e δ(v) x e b v v V e δ(v) x e b v e E x e {, d e } e E x e d e Figure 1: Integer Program and LP relaxation for the Demand Matching Problem min st S I y S S I:v S y S 1 y S v V (G), S I Proof of Lemma 2: Consider any instance of the demand matching roblem on a biartite grah G = (V, E) We use OPT for the maximum value of the demand-matching roblem and use OPT for the maximum value of the linear rogram We will bound the maximum ossible value of the integrality ga OP T /OP T We define an order on all the edges, such that any two edges e and e, either e > e or e > e, and e > e if d e > d e In the demand-matching roblem, given a subgrah H, let e H (v) be the largest edge incident to v We say a subgrah H is near b-bounded if for any vertex v V, we have d e b v + d eh (v) e δ H (v) We also call the edge e H (v) heavy edge for the vertex v Observe that there is exactly one heavy edge for each vertex while an edge can be heavy for both of its endoints Given a subgrah H, let the rofit of H be the sum of π e over all the edge of H, denoted by π(h) The following follows essentially from Sheherd and Vetta [SV7] Lemma 5 There is a near b-bounded subgrah H, such that π(h) OP T Let G be the subgrah of the line grah L(H) of H, such that two edges e, f E(H) are adjacent in G if they share a vertex in H and at least one of e, f is big for the common vertex v The following claim is also resent in Sheherd and Vetta [SV7] We include the roof for comleteness Claim 6 Any indeendent set of G corresonds to a feasible solution for demand matching roblem Proof Given any indeendent set A of G, it is an edge set in H and therefore G For any vertex v in H, among the edges incident to v in H, let e H (v) be a heavy edge Then either e H (v) is not in the indeendent set or only e H (v) in the indeendent set Since H is near b-bounded, in either case, e A δ(v) d e b v Thus A satisfies all the constraints in the demand matching roblem Claim 7 G is a tree-net 4

5 Proof For any cycle in G, suose x is the minimum vertex in the order of V (H) Let y and z be the two neighbors of x in the cycle Due to the minimality of x, it is not big for the common vertex with y and the common vertex with z Hence it does not have any other neighbors in G That is x has degree 2 If y and z are neighbors, then since G is simle, the edges x, y and z must form a triangle in G This is a contradiction since G is biartite Thus G is a tree-net Since G is a tree net, the hyothesis in Lemma 2 imlies that there is a fractional coloring y using at most ρ colors Now samle an indeendent set X of G such that any indeendent y set I is samled with robability I χ f (G) Observe that each vertex of G is included in X with robability at least y I I:v I χ f (G) 1 χ f (G) 1 ρ Since X is indeendent, Claim 6 imlies that X is a feasible demand matching in G where each edge of G is in X with robability at least 1 ρ Thus E[π(X)] = E[ π e ] = π e P r[e X] 1 π e 1 ρ ρ OP T e X e E e E 22 Fractional Coloring and -nice Distributions In this section, we rove Lemma 3 First we give a few definitions Let G be a tree and µ be a distribution over the indeendent sets of G We say µ is -nice if it satisfies the following roerties 1 P r I µ [v I] = v V (G) 2 P r I µ [u / I, v / I] for any nonadjacent vertices u, v in V (G) We rove the following lemma which relates µ-nice distribution to fractional chromatic number of tree-nets The roof aears in the aendix Lemma 8 Let be such that for every tree T there exists a -nice distribution µ over the indeendent sets of T Then the fractional chromatic number of any tree-net is at most 1 In this section, we construct a -nice distribution which then gives an aroximation algorithm roving Theorem 1 Proof of Lemma 3: Let = 1/27852 We will give a randomized algorithm to generate a indeendent set S of any tree T, such that P r(v S) = Also, if u and v are two non-adjacent vertices of T, we have P r(u, v S) Thus we will show that there is a -nice distribution over the indeendent sets of T Lemma 8 will then comlete the roof The algorithm will generate a indeendent set in a to down manner and maintain an inductive hyothesis We first state the inductive hyothesis and then give the algorithm which achieves this induction hyothesis For technical reasons, we attach a ath r 1 r 2 r 5 to T, such that r 5 V (T ) Let r = r 1 be the other endoint of the ath Let T be the new tree We consider T as a rooted tree with root r Let x, x 1, x 2, x 3, x 4, x 5 be constants such that = x = x 1 x 2 x 3 x 4 x 5 = and x 1 +x 2 +x 3 +x 4 +x 5 = 1 We are going to determine the value of x 2, x 3, x 4 later Given a vertex v Let U(v) be its arent vertex in T if it exists Let U (v) = v and U (v) = U(U 1 (v)) for any 1 be the th ancestor of v We are going to randomly generate an indeendent set S which satisfies the following conditions: 5

6 1 For any in {, 1, 2, 3, 4}, P r(u (v) S, U i (v) S i 1) = x for any vertex v T for which U (v) exists 2 For any nonadjacent vertices u and v in T, P r(u, v S) By condition (1), for =, we have P r(v S) = for each v V By condition (1), when = 1, ensures P r(u(v) S, v / S) = Thus P r(u(v) S, v S) = and this guarantees that S is a indeendent set The events that {U (v) S, U i (v) S i 1} for = {, 1, 2, 3, 4} are disjoint subevents within {v S} with total robability x +x 1 +x 2 +x 3 +x 4 = 1 But the event {U i (v) S i {, 1, 2, 3, 4}} is disjoint from all these event Hence we have another condition 3 P r(u i (v) S i {, 1, 2, 3, 4}) = The condition (2) will guarantee that the distribution is -nice which will finish the roof To fulfill all these conditions, we describe the algorithm to generate our indeendent set S We initialize S to be the emty set The algorithm starts from the root r and decides inductively whether to include a vertex in S or not The decision for v will only deend on the least ancestor of v which is in S Observe that U (v) S for some 1 5 For v, we include it with the following rule for each 1 5 P r(v S U (v) S, U i (v) S 1 i 1) = 1 x x 1 (1) To formalize it, we need to ensure that the rocess is defined for r i, 1 i 5 who have less than five ancestors, which we now describe We first add r into S with robability After we mae the choice for r 1 = r, we add r 2 into S deends on whether r 1 is in S Case 1 if r 1 S, we do not add r 2 into S with robability x 1 /P r(r S) = 1 Case 2 if r 1 S, we do not add r 2 into S with robability (x 2 + x 3 + x 4 )/P r(r S) Thus we have P r(r 2 S) = x 1 + x 2 + x 3 + x 4 = 1 For l = {3, 4, 5}, assume that we have already decided the choice of r 1, r 2,, r l 1 Case 1: Suose there exist a, such that U (r i ) = r i S and r i +1,, r i 1 S We do not add r i into S with robability x /x 1 Case 2: Suose r 1, r i 1 S, we do not add r i into S with robability (x i + + x 4 )/P r(r 1,, r i 1 S) Observe that this ensures that when l = 5, we always add r 5 into S if none of its ancestors have been added In such a way, we can guarantee there is at lease one element of {r 1,, r 5 } S Now for any vertex v V (T ) other than r 1,, r 5, after all decisions for the ancestors of v have been taen to add them into S or not, there exists a unique, such that U (v) S, U i (v) S 1 i 1 We do not include v into S with robability x /x 1 P r(v S U (v) S, U i (v) S 1 i 1) = 1 x x 1 (2) In this algorithm, it is easy to verify that the conditions (1),(3) are satisfied and is done in the aendix Lemma 9 Conditions (1) and (3) are satisfied by for any choice of x 1,, x 5 such = x 1 x 2 x 5 = and 5 i=1 x i = 1 6

7 Now we are going to choose x 2, x 3, x 4 to guarantee condition (2) By condition (1), we have for any in {2, 3, 4, 5}, P r(v, U (v) S, U i (v) S 1 i 1) = P r(u (v) S, U i (v) S 1 i 1) P r(u (v) S, U i (v) S i 1) = x 1 x for any vertex v T for which U (v) exists We define a function d : V (T ) V (T ) N 2, such that d(x, y) = (a, b), where a, b are the unique integers such that U a (x) = U b (y) The following claim shows that the robability both x, y S deends only on the distance of x and y to their least common ancestor The roof is based on a simle case analysis and induction Lemma 1 For any x, y T, we have that P r(x, y S) = t(a, b) where (a, b) = d(x, y) and t is the function with following roerties 1 t(a, b) = t(b, a) for all a, b 2 t(, ) = 3 t(, 1) = x 1 = 4 t(, 2) = x 1 x 2 5 t(, 3) = x 2 x 3 6 t(, 4) = x 3 x 4 + ( x2)2 7 t(1, 1) = (x1 x2)2 x 1 + (x2 x3)2 x 2 8 t(1, 2) = (x1 x2)(x2 x3) x 1 9 t(2, 2) = (x1 x2)2 + (x2 x3)2 x 1 + (x3 x4)2 x 3 + (x2 x3)(x3 x4) x 2 + (x3 x4)2 x 2 + (x4 x5)2 x 4 + (x3 x4)(x4 x5) x 3 + (x4 x5)2 x 3 1 If b = 3 and a 2, then we have t(a, b) = x1 x2 x 4 x 5 x 2 (x a+1 x a+2 ) 11 If b = a = 3 then we have t(a, b) = x1 x2 x 4 x 5 x 2 ( (x1 x2)(x2 x3) + x 4 x 5 ) 12 If b = 4, we have t(a, 4) = x1 x2 13 If b 5 then t(a, b) = x1 x2 t(a, 2) + x2 x3 t(a, b 2)+ x2 x3 t(a, 1) + x2 x3 t(a, ) + x3 x4 t(a + 1, ) + t(a, 1) + x2 x3 t(a, ) + x3 x4 t(a + 1, ) + t(a, 1) + x3 x4 t(a, ) + x4 x5 t(a + 1, ) t(a, b 3)+ x3 x4 t(a, b 4)+ x4 x5 t(a, b 5) It is easy to chec that the above conditions define t(a, b) for each a, b Now we are going to get the value for x 1,, x 5, where = x 1 x 2 x 5 =, 5 i=1 x i = 1 We just need P r(x, y S) = t(d(x, y)) 3 1 for any nonadjacent vertices x, y V (T ) Secially, we need to have t(1, 2) = t(2, 1) 3 1 That is (x 1 x 2 )(x 2 x 3 ) x 1 + (x 2 x 3 )(x 3 x 4 ) x 3 + (x 3 x 4 )(x 4 x 5 ) x When = 1/27852, there is a solution of x = (x 1,, x 5 ) satisfies the constraint That is x = (3692, 257, 516, 43, ) 7

8 With this assignment of x, we get all the value of t(a, b) for a, b 5 as following: and S = (t(a, ) : 1 a 5) = (, 1635, 1541, 1197, 148) T = (t(a, b) : 1 a, b 5) = We have t(a, b) 3 1 = 176 whenever i and j are not more that 5 When b 6, t(a, b) is a convex combination of t(a, b 2),, t(a, b 5) By induction, we always have t(a, b) 3 1 From the above discussion, we give a -nice distribution with = 1/27852, hence we conclude that the integrality ga is at most Here we assume x 5 =, the robability of each vertex being included in S deends on the choice of at most its five lowest ancestors Actually, if we consider one more ancestor in the deendency, and assume x 1 x 5 x 6 = instead, we still see the maximum value of t(1, 2) is obtained when x 5 = Hence we conjecture that the uer bound can not be imroved 3 Lowerbound In this section, we are going to give a family of demand matching roblems, indexed by integer, such that the integrality ga is at least 2699 when is aroaching to infinity Let T be a rooted tree with root r, such that r has only one neighbor s, and after deleting r, T is a -ary tree with root s Given a vertex v in T, we say the level of v, denoted by l(x), is the distance to r For examle, l(r) = and l(s) = 1 For x r, let e x be the edge between x and the arent of x For the vertex x at level l, We let the caacity of vertex x to be b x = 2l For the edge e between level l 1 and l, let the demand of edge e be d e = 2l Thus for any x at level l, there is one incident edge e x with demand 21 and all other incident edges have demand 2(l+1) For any two vertices x and y which have distance (2, 1), we add an edge xy to T with demand d xy being 4 We call such an edge a small edge for both x and y We called the above define grah G We consider the demand matching roblem with the giving demand function d on the edges of G and the constraint function b on the vertices of G For every vertex x at level l, there are edges incident to it with demand 2(l+1), there are totally at most 1 + ( 1) (= 2 ) edges incident to it with demand 2 3 So d e 2l + 2l = 2l + 2l l + 2l = b(v)+d e(v) e δ G (x) Thus e x is a big edge for x G itself is near b-bounded Since d ex = b v = 2l, a subgrah M is a demand matching, if and only if it does not include e x or it does not contains any other edge incident to x, for each non-root vertex x Also, Since e δ G (x) d e b(v) 2l + 2 2l 1 2l 8 = 2 +,

9 So x e = +2 d e for all edge e E(G ) is a solution for the linear rogramming Following the argument for uer bound, we define H be the subgrah of the line grah L(G ), such that any two edges are adjacent if and only if they share a vertex x and one of them is big to x, which is e x It is easy to verify that an edge set A in H induces a demand matching if and only if it is a indeendent set in H Hence we have The following lemma follows from duality χ f (G ) = χ f (H ) Lemma 11 There exists a rofit function such that π(f ) 1 for each feasible demand matching F and the otimal LP solution has rofit at least +2 χ f (H ) This imlies that the integrality ga is at least +2 χ f (H ) for any Note that H is a tree-net obtained from a -ary, level tree T by adding a vertex adjacent to any two vertices on T with distance (2, 1) We are going to rove that the fractional chromatic number lim χ f (H ) 2699, which imlies that integrality ga max OP T OP T lim + 2 χ f (H ) 2699 Thus to rove the Theorem 4, we just need to rove the following lemma Lemma 12 Let H is the tree-net obtained from a -ary, level tree T by adding a vertex adjacent to any two vertices on T with distance (2, 1) We have lim su χ f (H ) > 2699 Proof Suose χ f (H ) = 1/ 2699 for all Let I be the be collection of all indeendent sets of H Since χ f (H ) = 1/, there exists a distribution µ of I, such that P r I µ (v I) for any v V (H ) This equivalent to the following two conditions: (1) P r I µ (v I) for any vertex v V (T ); (2) P r I µ (u, v I) for any air of vertices u, v with distance (2,1) in T The condition (2) is equivalent to the following statement: (3) P r i µ (u, v I) 3 1 for any air of vertices u, v with distance (2,1) in T Assume µ is a maing { of [, 1] to I We obtain a maing g from V (G) [, 1] to {, 1}, 1, x µ(t) such that g(x, t) =, x µ(t) Proosition 13 g is a function that has the following roerties: 1 g(x, t)dt = x V (T ); 2 g(u, t)g(v, t) = for any adjacent vertices u, v in V (T ) and t [, 1]; 3 (1 g(u, t))(1 g(v, t))dt for any nonadjacent vertices u, v in V (T ); 4 g(u, t)g(v, t)dt 3 1 for any nonadjacent vertices u, v in V (T ) Let x and y be two non-leaf vertices in T such that y is a child of x in T Let u 1,, u be all the children of x where u = y, and let v 1,, v be all the children of y For any i [1, 1] and j [1, ], u i and v j are not adjacent Therefore we have = 1 i=1 g(u i, t)g(v j, t)dt 3 1, ( g(u i, t)) ( g(v i, t))dt ( 1)(3 1) j=1 i [1, 1], j [1, ] 9

10 Since g(u, t) ( j=1 g(v i, t))dt =, we have that i=1 g(u i, t) j=1 g(v i, t) dt 1 (3 1) Given a non-leaf vertex w in T and any t [, 1], let h(w, t) = v is a child of w g(v, t)/ be the ratio of children of w iced in the indeendent set Then Since h(w, t) = h(w, t) (1 g(w, t)), we have h(x, t)h(y, t)dt 1 (3 1) h(x, t) (1 g(y, t)) h(y, t) (1 g(x, t))dt 1 (3 1) By Cauchy-Schwartz inequality, we have ( 1 (3 1))2 ( = h(x, t)(1 g(y, t)) h(y, t)(1 g(x, t))dt) 2 h(x, t) 2 (1 g(y, t)) 2 dt h(x, t) 2 (1 g(y, t))dt h(y, t) 2 (1 g(x, t)) 2 dt h(y, t) 2 (1 g(x, t))dt where the last equality uses the fact that g is, 1-valued Since h(x, t)2 (1 g(y, t))dt 1 and h(y, t)2 (1 g(x, t))dt 1, we have ( 1 (3 1))2 h(x, t) 2 (1 g(y, t))dt < 1 Let l(v) = h(v, t)2 (1 g(u, t))dt, where v is any non-root vertex and u is the arent of v in T We have ( 1 (3 1))2 < l(v) < 1 (3) In the next claim we identify the vertex x with nearly the maximum value of l(v) over all vertices v Since we do not want it to be a leaf vertex or root, we can claim that it is nearly the maximum and not exactly maximum Claim 14 There exist a function a() which aroaches 1 as aroaches infinity and a vertex x which is not the root and not a leaf in T, such that l(x) a()l(y) for any child y of x Proof Let a() = ( 1 (3 1))2/( 2) Suose the claim is not true Then there is a ath P = x 2 x 3 x 1, such that x 2 is a child of the root, x i+1 is a child of x i for i = 2,, 2, such that l(x i ) < a()l(x i+1 ) for i = 2,, 2 Thus, we have l(x 2 ) < l(x 1 )a() 2 l(x 2 )( 1 (3 1))2 < ( 1 (3 1))2 since l(x 2 ) < 1 But this contradicts equation (3) 1

11 Assume x is the vertex in the claim, let z is its arent Let y be any of its child We have ( 1 (3 1))2 = h(x, t) 2 (1 g(y, t))dt h(x, t) 2 (1 g(y, t))dt l(y) h(x, t) 2 (1 g(y, t))dt l(x) a() h(x, t) 2 (1 g(y, t))dt h(y, t) 2 (1 g(x, t))dt h(x, t) 2 1 (1 g(z, t))dt a() Since it is true for any child y of x, we have a()( 1 (3 1))2 = h(x, t) 2 1 (1 g(y, t))dt y is a child of x h(x, t) 2 (1 h(x, t))dt h(x, t) 2 (1 g(z, t))dt h(x, t) 2 (1 g(z, t))dt We now show that it is a contradiction when = Claim 15 h(x, t)2 (1 h(x, t))dt h(x, t)2 (1 g(z, t))dt C when = , where C = 1235 < (3 1) 2 Now we now that if = 1/2699, then h(x, t)2 (1 h(x, t))dt h(x, t)2 (1 g(z, t))dt C < (3 1) 2 However, if G has fractional chromatic number at most 2699 for any 1, we will have a()( 1 (3 1))2 when is big enough, we will have h(x, t) 2 (1 h(x, t))dt h(x, t) 2 (1 g(z, t))dt h(x, t) 2 (1 h(x, t))dt Contradiction! This comletes the roof of the lemma h(x, t) 2 (1 g(z, t))dt a()( 1 (3 1))2 > C References [CG7] [CK5] [Clo53] José R Correa and Michel X Goemans Imroved bounds on nonblocing 3-stage clos networs SIAM J Comut, 37(3):87 894, 27 Chandra Cheuri and Sanjeev Khanna A olynomial time aroximation scheme for the multile nasac roblem SIAM J Comut, 35(3): , 25 Charles Clos A study of non-blocing switching networs Bell System Technical Journal, 2:46424,

12 [DGG99] Yefim Dinitz, Naveen Garg, and Michel X Goemans On the single-source unslittable flow roblem Combinatorica, 19(1):17 41, 1999 [DGHK99] D Z Du, B Gao, F K Hwang, and J H Kim On multirate rearrangeable clos networs SIAM J Comut, 28(2):463 47, February 1999 [Eug31] [FS8] [Kle96] Egerváry Eugene Matrixo ombinatorius tulajdons/ agairöl, [in Hungarian: On combinatorial roerties of matrices] Matematiaiés Fiziai Lao, 38:16 28, 1931 Uriel Feige and Mohit Singh Edge coloring and decomositions of weighted grahs In Dan Halerin and Kurt Mehlhorn, editors, Euroean Symosium on Algorithms, volume 5193 of Lecture Notes in Comuter Science, ages Sringer, 28 Jon M Kleinberg Single-source unslittable flow In Annual IEEE Symosium on Foundations of Comuter Science, ages 68 77, 1996 [Kön31] D König Grahoès matrixo, [in Hungarian: Grahs and matrices] Matematiaiés Fiziai Lao, 38: , 1931 [KS1] [Kuh55] [NV3] [Par11] [Su] [ST93] [SV7] Aendix Stavros G Kolliooulos and Clifford Stein Aroximation algorithms for singlesource unslittable flow SIAM J Comut, 31(3): , 21 H W Kuhn The Hungarian method for the assignment roblem Naval Research Logistics Quarterly, 2(1-2):83 97, 1955 Hung Q Ngo and Van H Vu Multirate rearrangeable clos networs and a generalized edge coloring roblem on biartite grahs In Annual ACM-SIAM Symosium on Discrete Algorithms, ages ACM/SIAM, 23 Ojas Pareh Iterative acing for demand and hyergrah matching In Integer Programming and Combinatorial Otimization Conference, ages , 211 Martin Sutella Aroximating the single source unslittable min-cost flow roblem In Annual IEEE Symosium on Foundations of Comuter Science, ages IEEE Comuter Society, 2 David B Shmoys and Éva Tardos An aroximation algorithm for the generalized assignment roblem Math Program, 62: , 1993 F Bruce Sheherd and Adrian Vetta The demand-matching roblem Math Oer Res, 32(3): , 27 Proof of Lemma 5: Given a feasible solution x to the linear rogramming, we say an edge e is an fraction edge if < x e < d e Let F = F x (G) be the subgrah of G induced by the factional edges Assume x is a feasible solution with minimum number of fractional edges, such that e δ G (v) x e b (v) We claim that there is no cycle in F x(g) Suose that there is a cycle with edges e 1,, e 2 in order, let t be a vector such that t e2i 1 = 1, t e2i = 1 and t e = for any other edges in G Since e 1,, e 2 are fractional edges, then when ε > is small enough, we have x + εt and x εt both satisfies the require solution We can be Assume ε is the maximum such number, then one of F x +εt(g) and F x ε will have fewer edges comare to x Contradiction! Initially, let x = x Iteratively, we are going to change the value of x to reduce the number of fractional edges, such that e E π ex e does not decrease Suose F x (G) is not trivial and 12

13 has no cycle Let v and u be two leaves lie on the same comonent in F x (G) There is a unique vu-ath with edges e 1,, e such that e 1 incidents v and e incidents u Let t be vector such that t e2i 1 = 1, t e2i = 1 for edges in {e 1,, e }and t e = for all other edges Suose e E π et e, there is a ε, such that x + εt satisfies the constraint, with less fractional edges, and x + εt has more rofit than x e We relace x by x + εt Suose e has negative rofit, then we relace x by x εe for some ε > After such iterative oerations, we finally will have F x (G) has no edges Let e v be one last edge in δ(v) that disaear from F x (G) For each vertex v V (G), in each ste, we either have that e δ(v) x e does not change, with only ossible excetion that for one ste that e v disaear from F x (G) as a leaf, and e δ(v) change by a number less than d e Since initially, we have e δ(v) x e b v, finally we have e δ(v) x e b v + e v b v + e(v) and x e be either or d e In addition, x has rofit at least OP T Let H be the grah induced by edges with nonzero value in x, we have H is near b-bounded, and π(h) OP T Proof of Lemma 8: Let G be a tree-net Let T be a maximal induced subtree of G The edges of G \ T can be decomosed as edge disjoint aths of length two joining non-adjacent vertices of T Letµ be a -nice distribution over T We will extend this distribution to indeendent sets of G and ensure that each vertex is included in the indeendent set with robability at least This will ensure that the fractional chromatic number of G is at most 1 For each indeendent set I in the suort of I, we extend it to include the vertices of G \ T Consider any ath P of length two joining non-adjacent vertices of T and w be the only internal vertex of the ath We will extend the distribution µ to w Let u, v be the two neighbors of w in T For any I in the suort of µ, whenever u, v / I, we extend I I w Since P r[u, v / I] thus w will be included with robability at least Vertices of T already are resent will robability at least since µ was -nice Proof of Lemma 9: A simle chec ensures that the conditions are satisfied for r 1,, r 5 This serves as base case for the induction Now, let v T be any other vertex Condition 1 By induction on roerty (4), we have that at least one of U i (v) S for some 1 i 5 Thus 5 P r(v S) = P r(v T U i (v) S, U j (v) / S 1 j i 1) P r(u i (v) S, U j (v) / S 1 j i 1) = i=1 5 (1 x i ) x i 1 = x x 5 = x i 1 i=1 where the first term follows from the algorithm and the second term follows from induction hyothesis Thus the condition is true for = Now for any 1 P r(u (v) S, U i (v) / S i 1) = = P r(u (v) S, U i (v) / S 1 i 1)P r(v / S U (v) S, U j (v) / S 1 j 1) = x x 1 x 1 = x 13

14 Condition 3 This follows simly if condition 1 is satisfied Proof of Lemma 1: First it is easy to see that if it is true, t(a, b) = t(b, a) We rove rest of the conditions by induction on a + b It is easy to verify that t(, ) = P r(x S) =, t(, 1) = t(1, ) = x 1 = roving (2) and (3) When d(x, y) = (, 2), then x = U 2 (y) Then P r(x, y S) = P r(u 2 (y) S, y S, U(y) S) = P r(u 2 (y) S, U(y) S) P r(u 2 (y) S, y / S, U(y) S) = x 1 x 2 Thus t(, 2) = t(2, ) = x 1 x 2 = x 2 When d(x, y) = (, 3), ie, x = U 3 (y), we have P r(x, y S) = P r(x, y S, U(y), U 2 (y) S) = P r(u 3 (y) S, U(y), U 2 (y) S) P r(u 3 (y) S, y, U(y), U 2 (y) S) = x 2 x 3 Therefore t(, 3) = t(3, ) = x 2 x 3 When d(x, y) = (, 4), ie, x = U 4 (y) We have P r(x, y S) = P r(y, U 4 (y) S, U(y), U 2 (y), U 3 (y) S) + + P r(y, U 2 (y), U 4 (y) S, U(y), U 3 (y) S) = x 3 x 4 + P r(y U 2 (y) S)P r(u 2 (y), U 4 (y) S, U 3 (y) S) = x 3 x 4 + (1 x 2 x 1 )(x 1 x 2 ) = x 3 x 4 + ( x 2) 2 Suose d(x, y) = (1, 1) and z = U(x) = U(y) We have P r(x, y S) = P r(x, y, U(z) S, z S) + P r(x, y, U 2 (z) S, z, U(z) S) We have +P r(x, y, U 3 (z) S, z, U(z), U(z) 2 S) + P r(x, y, U 4 (z) S, z, U(z), U 2 (z), U 3 (z) S) P r(x, y, U(z) S, z S) = P r(x, U(z) S, z S)P r(y, U(z) S, z S)/P r(u(z) S, z S) = (x 1 x 2 ) 2 /x 1 similarly, we get P r(x, y, U 2 (z) S, z, U(z) S) = (x 2 x 3 ) 2 /x 2, P r(x, y, U 3 (z) S, z, U(z), U 2 (z) S) = (x 3 x 4 ) 2 /x 3, and P r(x, y, U 4 (z) S, z, U(z), U 2 (z), U 3 (z) S) = (x 4 x 5 ) 2 /x 4, Thus we have t(1, 1) = P r(x, y S) = (x 1 x 2 ) 2 x 1 + (x 2 x 3 ) 2 x 2 + (x 3 x 4 ) 2 x 3 + (x 4 x 5 ) 2 x 4 Suose d(x, y) = (1, 2), let z = U(y) and w = U 2 (y) = U(x) Similar to the argument for 14

15 t(1, 1) We have t(1, 2) = t(2, 1) = P r(x, y S) = P r(x, y, U(w) S, z, w S) + P r(x, y, U 2 (w) S, z, w, U(w) S) + +P r(x, y, U 3 (w) S, z, w, U(w), U 2 (w) S) = (x 1 x 2 )(x 2 x 3 ) x 1 + (x 2 x 3 )(x 3 x 4 ) x 2 + (x 3 x 4 )(x 4 x 5 ) x 3 Suose d(x, y) = (2, 2) Let z = U 2 (x) = U 2 (y) We have t(2, 2) = P r(x, y S) = P r(x, y, z S, U(x), U(y) S) + P r(x, y, U(z) S, U(x), U(y), z S) +P r(x, y, U 2 (z) S and U(z), z S) + P r(x, y, U 3 (z) S, U 2 (z), U(z), z, U(x), U(y) S) = (x 1 x 2 ) 2 + (x 2 x 3 ) 2 x 1 + (x 3 x 4 ) 2 x 2 + (x 4 x 5 ) 2 x 3 Suose d(x, y) = (a, b), where b 5 We have t(a, b) = P r(x, y S) = P r(x, y, U 2 (y) S, U(y) S) + P r(x, y, U 3 (y) S, U(y), U 2 (y) S) +P r(x, y, U 4 (y) S, U(y), U 2 (y), U 3 (y) S) + P r(x, y, U 5 (y) S, U(y),, U 4 (y) S) = P r(y, U 2 (y) S, U(y) S) P r(u 2 (y) S) P r(x, U 2 (y) S) + P r(y, U 3 (y) S, U(y), U 2 (y) S) P r(u 3 P r(x, U 3 (y) S) (y) S) + P r(y, U 4 (y) S, U(y), U 2 (y), U 3 (y) S) P r(u 4 P r(x, U 4 (y) S) (y) S) + P r(y, U 5 (y) S, U(y),, U 4 (y) S) P r(u 5 P r(x, U 5 (y) S) (y) S) = x 1 x 2 t(a, b 2) + x 2 x 3 t(a, b 3) + x 3 x 4 t(a, b 4) + x 4 x 5 t(a, b 5) When b = 4, the only difference is that we can not have t(a, 1) By a slight adjustment, we have t(a, 4) = x 1 x 2 t(a, 2) + x 2 x 3 t(a, 1) + x 3 x 4 t(a, ) + x 4 x 5 t(a + 1, ) Similarly, when b = 3, let z = U a (x) = U 3 (y), we should be careful for the last term: t(a, 3) = x 1 x 2 t(a, 1) + x 2 x 3 t(a, ) + x 3 x 4 t(a + 1, ) + P r(y, U 2 (z) S, U(y), U 2 (y), z, U(z) S) P r(u 2 P r(u 2 (z), x S, z, U(z) S) (z) S, z, U(z) S) where P r(y, U 2 (z) S, U(y), U 2 (y), z, U(z) S) P r(u 2 (z) S, z, U(z) S) = x 4 x 5 x 2 15

16 and when a 2, we have P r(u 2 (z), x S, z, U(z) S = x a+1 x a+2 when a = 3 P r(u 2 (z), x S, z, U(z) S) = P r(u 2 (z), x, U 2 (x) S, U(x), z, U(z) S) + P r(u 2 (z), x S, U(x), U 2 (x), z, U(z) S) = (x 2 x 3 )(x 1 x 2 )/ + x 4 x 5 From here, it is easy to get that P r(x, y S) only deends on d(x, y) for all d(x, y) Proof of 11: For the grah H, consider the following linear rogram where D denotes the set of all indeendent sets The otimal solution is of value χ f (H ) min st F D y F F D:e F d ey F 1 e V (H ), y F Also consider the dual of the linear rogram max st e V (H ) z e F D e V (H ) z e 1 F D, z e e V (H ) By duality, there exists a otimal solution z of objective value χ f (H ) Let z be the rofit function The constraints of the dual imly that every indeendent set of H which corresonds to a demand matching of G has rofit at most one Moreover the objective value is exactly e E(G ) z e = χ f (H ) But then the linear rogramming solution x E = +2 has rofit at least +2 χ f (G) Proof of Claim 15: We have g(x, t)dt = g(z, t)dt = g(x,t)=1 g(z,t)=1 1dt = (4) 1dt = (5) h(x, t) = v is a child of g(v, t)/ is a ste function with ossible values, 1/, 2/,, 1 Let x A i = {t : g(z, t) = 1, h(x, t) = i } and let B i = {t : g(z, t) =, h(x, t) = i/} for i =, 1,, Let a i = A i be the measure of A i and let b i = B i be the measure of B i We have the following roerties: (1) i= i (a i + b i ) = h(x, t)dt =, (2) i= a i = g(z, t)dt =, (3) i= b i = (1 g(x, t))(1 g(z, t))dt =

17 and Let A = h(x, t)2 (1 h(x, t))dt, and let B = h(x, t)2 (1 g(z, t))dt We have A = (( i )2 ( i )3 )(a i + b i ) i= B = ( i )2 b i We are going to rove that given the condition (1), (2), (3), we have Claim 16 Given the condition that i= max AB C < (1 3) 2, when = 1/2699 i= i (a i + b i ) =, a i =, i= b i = 1 2 i= and a i, b i, to maximize AB = ( (( i )2 ( i )3 )(a i + b i ))( ( i )2 b i ), i= there are at most three of a 1, a and three of b 1, b which are nonzero Proof Consider A, B as functions of a,, a and b,, b Let L = AB λ( i= i (a i + b i ) ) β( i= a i ) ω( i= b i (1 2)) By Lagrange Multiliers, to maximize AB, if a i for some i [, ], we have L =, a i that is i= [( i )2 ( i )3 ]B λ i β = There are at most 3 of i [, ] satisfies the equality, which imlies that there are at most 3 of a,, a which are nonzero Similarly, if b i for some i [, ], we have L b i =, that is [( i )2 ( i )3 ]B + A( i )2 λ i ω = There are at most 3 of i [, ] satisfies the equality, which imlies that there are at most 3 of b,, b which are nonzero 17

18 Now assume x 1, x 2, x 3 be the three ossible value of i/ for which a i is nonzero, and let a 1, a 2, a 3 be the corresonding value of a i Let y 1, y 2, y 3 be the three ossible value of i/ for which b i is nonzero, and let b 1, b 2, b 3 be the ossible value of b i Using Matlab, we can rove that under the constraint that: (1) a 1 + a 2 + a 3 = ; (2) b 1 + b 2 + b 3 = 1 2 ; (3) a 1x 1 + a 2x 2 + a 3x 3 + b 1y 1 + b 2y 2 + b 3y 3 = ; (4) a i, b i, x i, y i 1 for i = 1, 2, 3, we have max AB C = 1234 < 1243 = (3 1) 2, when = 2699, where A = 3 i=1 (x2 i x3 i )a i + (y2 i y3 i )b i and B = 3 i=1 y2 i b i By the above claim, we always have h(t) 2 (1 h(t))dt h(t) 2 dt C < (3 1) 2, when = 1/